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On the doubly infinite ladder graph, self-avoiding walks from a fixed graph vertex using only rightward or vertical steps are in bijection with words in $R,V$ containing no consecutive $V$ symbols. Their count is ${\sigma }_n=F_{n+2}$, since ${\sigma }_0=1$, ${\sigma }_1=2$, and ${\sigma }_n={\sigma }_{n-1}+{\sigma }_{n-2}$. Its exponential growth rate is the golden ratio.

The doubly infinite ladder graph has graph vertices $\mathbb{Z}\times \{0,1\}$, horizontal edges $\{(j,s),(j+1,s)\}$, and vertical edges $\{(j,0),(j,1)\}$.

A ladder graph is the Cartesian product of graphs of a graph path with the two-vertex complete graph $K_2$. Its two parallel rails are joined by a rung at each position. The doubly infinite ladder graph uses an infinite rail indexed by $\mathbb{Z}$.

Represent the doubly infinite ladder graph by $\mathbb{Z}\times \{0,1\}$, with horizontal edges between successive columns and a vertical rung at each column. An allowed self-avoiding walk is coded by a word in $R,V$. Two successive $V$ steps revisit the preceding graph vertex and are forbidden. Conversely, every word without consecutive $V$ steps is a self-avoiding walk: horizontal steps strictly increase the column, and a column is visited vertically at most once. This establishes a bijection, not merely an upper bound.

For the directed ladder self-avoiding walk count, ${\sigma }_0=1$, ${\sigma }_1=2$, and for $n\ge 2$ a word either ends in $R$, or ends in $RV$. Removing that final block givesLet $\phi =(1+\sqrt{5})/2$ be the golden ratio, and $\psi =(1-\sqrt{5})/2$. The Binet formula for the Fibonacci number yieldsSince $|\psi |<\phi$, taking $n$th roots gives ${\lim }_{n\to \infty }{\sigma }_n^{1/n}=\phi$. This use of the golden-ratio symbol is independent of the connection rate in Question 1.